StGeorgeDragons
Member
- Joined
- Aug 13, 2004
- Messages
- 527
the letters of the word BANANAS are written on separate cards and drawn at random from a hat. calculate the probability that : an A will be drawn first or last but not both
klaw said:P(A drawn first or last)= P(A drawn first but not last)+P(A drawn last but not first)=3/7*2/3+4/7*1/2=4/7
sorry wrong....currysauce said:i'll have a crack at it, even tho i am hopeless at perms
overall probability = 7! / 3!2!
Lets see, umm If you have A at the start, then 5 other positions can be filled 5!/2!2! then you x 2? for the reverse, if the A is at the end (i really got no clue)
therefore
5!/2!2! // 7!/3!2! = 30/420 = 3/42 = 1/14?
azn_gangsta81 said:do you have the answer for it coz i wanna see if i got it right before posting the steps
can't be, that means there's a 686% chance of happeningsue_ad said:Is It 48/7 ?